Prove $$\frac{1}{2}^x+\frac{1}{2}^\frac {1}{x}\leq 1$$, where $x $ is a positive real number. This problem is from my friend. Here is my approach. It is sufficient to show that $$0\leq2^x-1-2^{x-\frac {1}{x}}$$. To find minimum of right side, I differentiate the right fuction, but I cannot find all zeros of $2^t-1-t^2$, where $t=\frac{1}{x}$. Do you have any idea to find upperbound?

  • $\begingroup$ They meet at t=0,1 and one more point between 4 and 5. $\endgroup$ – user148928 Jun 9 '15 at 2:58
  • $\begingroup$ The problem is it takes a minimum at a point between 4 and 5. $\endgroup$ – user148928 Jun 9 '15 at 3:19
  • $\begingroup$ You need to worry only about $x\in (0,1)$, the LHS is invariant to $t \mapsto \frac1t$. $\endgroup$ – Macavity Jun 9 '15 at 3:21
  • $\begingroup$ take $f(t)=ln2*t-ln(1+t^2)$,prove on(0,1) there is only one max positive point, so two ends will be min. , $\endgroup$ – chenbai Jun 9 '15 at 3:29
  • $\begingroup$ @Macavity Thank you for your comment. $\endgroup$ – user148928 Jun 9 '15 at 4:12

Considering the function $$f(x)=1-2^{-1/x}-2^{-x}$$ As Macavity commented, $f(\frac1x)=f(x)$ and you only need to consider $x\in (0,1)$. Look at the derivatives $$f'(x)=2^{-x} \log (2)-\frac{2^{-1/x} \log (2)}{x^2}$$ $$f''(x)=-\frac{2^{-1/x} \log ^2(2)}{x^4}+\frac{2^{1-\frac{1}{x}} \log (2)}{x^3}-2^{-x} \log ^2(2)$$ Using limits for the lower bound $x=0$, you can show that $$f(0)=0 \\\ f'(0)=\log (2) \\\ f''(0)=-\log ^2(2)$$ Similarly $$f(1)=0 \\\ f'(1)=0 \\\ f''(1)=\log (2)-\log ^2(2)$$ Looking at the first derivative, it cancels close to $x=0.2$; using Newton method starting at this point, the successive iterates are $0.214110$, $0.215101$, $0.215106$ which is the solution for six significant figures. At this point, the second derivative is $-3.80712$ which confirms that the point is a maximum.

So, for the considered interval, $f(x)\geq 0$ and it reaches a maximum value $\approx 0.0986559$ at $x\approx0.215106$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.